Physics 5300, Theoretical Mechanics Spring 2015
Assignment 5 solutions
The problems numbers below are from Classical Mechanics, John R. Taylor, University
Science Books (2005).
Problem 1
Solution:
Taylor 10.10
(a)
M
I=
L
Z
M
I=2
L
Z
L
dxx2 =
M L3
M L2
=
L 3
3
(1)
x2 dx =
M L2
2M L3
=
L 24
12
(2)
x=0
(b)
Problem 2
L
2
x=0
Taylor 10.12
Solution: We first compute I for rotation about an axis parallel to the z axis, passing
through one corner. We will then use the parallel axis theorem to get I around an axis
passing through the centroid.
Let the length of the prism be L. The area of each side is 2a. The height is
p
√
h = 4a2 − a2 = 3a
(3)
The area is
√
√
1
A = (2a) 3a = 3a2
2
The volume is then
V =
√
3a2 L
(4)
(5)
and the density is
ρ=
M
M
=√
V
3a2 L
(6)
Let the y axis be along the perpendicular of the triangle, with y = 0 being the vertex of
the triangle. The range of y is
√
0 ≤ y ≤ 3a
(7)
1
At any given value of y, the range of x is
1
1
−√ y ≤ x ≤ √ y
3
3
(8)
The distance squared from the z axis is x2 + y 2 . Thus
Z
√
L
I = 2ρ
Z
3a
dz
Z
dy
z=0
y=0
1
√
y
3
dx(x2 + y 2 )
(9)
x=0
The first part is
Z
√
L
Z
I1 = 2ρ
Z
dy
dz
y=0
z=0
√
3a
√1 y
3
√
Z
2
3a
dxx = 2ρL
x=0
dy
y=0
x3 x= √13 y
|
3 x=0
(10)
y3
1 y 4 √3a
1
M
1
1 9a4
1
dy √ = 2ρL √
La4 = M a2
|y=0 = 2ρL √
= √ ρLa4 = √ √
2
6
9 3
9 3 4
9 3 4
2 3
2 3 3a L
y=0
(11)
The second part is
Z
3a
= 2ρL
Z
√
L
I2 = 2ρ
Z
3a
dz
z=0
dyy
y=0
2
Z
√1 y
3
Z
dx = 2ρ
x=0
√
L
Z
3a
dz
z=0
y=0
x= √1 y
dyy 2 x|x=0 3
(12)
√
√
3 3
3 3 M
1
2ρL y 4 √3a 2ρL 9a4
3
4
√
= 2ρ
dz
|y=0 = √
=
ρLa =
dyy √ = √
La4 = M a2
2
2
2
3
3 4
3 4
3a2 L
z=0
y=0
(13)
The centroid is at a distance
2√
2
(14)
3a = √ a
3
3
Z
L
√
Z
3a
3
from the vertex. Thus we need to subtract an amount
4
M a2
3
(15)
from the sum of the above two terms. This gives
1
4
1
3
I = M a2 + M a2 − M a2 = M a2
2
6
3
3
(16)
The products of inertia will vanish by symmetry, if we compute them around the
midpoints of the prism.
Problem 3
Taylor 10.15
2
Solution: (a) Let the axis of rotation be the z axis, and the cube extend from 0 to a
on each axis. The density is
M
ρ= 3
(17)
a
We get
Z
Z
Z
a
a
a
z=0
dx(x2 + y 2 )
dy
dz
I=ρ
(18)
x=0
y=0
We have
Z
a
Z
a
y=0
z=0
a
2
Z
a
Z
x=0
a
dy
dz
dxx = ρ
dy
dz
ρ
Z
y=0
z=0
a3
a5
M a2
a3
= ρa2
=ρ =
3
3
3
3
(19)
The other integral gives the same result, so we get
I=
(b) The height of the center of mass is
mass is at a height a2 . Thus the PE lost is
2M a2
3
√a
2
(20)
above the table. At the end, the center of
√
2−1
1
1
M ga( √ − ) = M ga
2
2 2
(21)
1 2
Iω
2
(22)
The KE gained is
Thus we get
Problem 4
Solution:
√
1 2M a2 2
2−1
ω = M ga
2 3
2
√
g3
ω2 =
( 2 − 1)
a2
(23)
(24)
Taylor 10.21
If we have a diagonal term like i = x, j = x then we get
Z
Z
2
2
2
2
Ixx = ρ(x + y + z − x ) = ρ(y 2 + z 2 )
which is correct. If we have an off diagonal term, then we get
Z
Ixy = ρ(−xy)
3
(25)
(26)
which is correct as well.
Problem 5
Solution:
Taylor 10.35
We have
X
Ixx =
mi (y 2 + z 2 ) = 2m[a2 + a2 ] + 3m[a2 + a2 ] = 10ma2
(27)
i
Iyy =
X
mi (x2 + z 2 ) = m[a2 ] + 2m[a2 ] + 3m[a2 ] = 6ma2
(28)
mi (x2 + y 2 ) = m[a2 ] + 2m[a2 ] + 3m[a2 ] = 6ma2
(29)
i
Izz =
X
i
Ixy = −
X
mi xy = m[0] = 0
(30)
mi yz = −m[0] − 2m[a2 ] − 3m[−a2 ] = ma2
(31)
i
Iyz = −
X
i
Ixz = −
X
mi xz = −m[0] − 2m[0] = 0
(32)
i
Thus


10 0 0
I = ma2  0 6 1 
0 1 6
The eigenvalues are found from the equation (keeping apart a factor of ma2 )


10 − λ
0
0
Det  0
6−λ
1 =0
0
1
6−λ
(33)
(34)
(10 − λ)[(6 − λ)2 − 1] = 0
(35)
λ = 10
(36)
One solution is
The other solutions are given by
(6 − λ)2 = 1,
6 − λ = ±1,
λ = 5, λ = 7
(37)
Thus the moments of inertia are
I1 = 10ma2 ,
I2 = 5ma2 ,
4
I3 = 7ma2
(38)
The eigenvectors are given by
 
 

Vx
Vx
10 0 0
 0 6 1   Vy  = 10  Vy 
Vz
Vz
0 1 6
(39)
Let Vx = 1. Then we get
6Vy + Vz = 10Vy ,
− 4Vy + Vz = 0,
Vy =
Vz
4
(40)
We also get
Vy + 6Vz = 10Vz ,
Vy − 4Vz = 0
(41)
Thus we get the eigenvector
(1, 0, 0)
(42)
 
 
10 0 0
Vx
Vx
 0 6 1   Vy  = 5  Vy 
0 1 6
Vz
Vz
(43)
For λ = 5 we get

Thus
10Vx = 5Vx ,
Vx = 0
(44)
6Vy + Vz = 5Vy ,
Vy = −Vz
(45)
Vy + 6Vz = 5Vz ,
Vy = −Vz
(46)
Thus we can take the eigenvector
1
1
(0, √ , − √ )
2
2
(47)
 
 
10 0 0
Vx
Vx
 0 6 1   Vy  = 7  Vy 
0 1 6
Vz
Vz
(48)
For λ = 7 we get

Thus
10Vx = 7Vx ,
Vx = 0
(49)
6Vy + Vz = 7Vy ,
Vy = Vz
(50)
Vy + 6Vz = 7Vz ,
Vy = Vz
(51)
Thus we can take the eigenvector
1 1
(0, √ , √ )
2 2
5
(52)
Problem 6
Solution:
Taylor 10.40
(a) We have
I1 ω̇1 − (I2 − I3 )ω2 ω3 = 0
(53)
L1 I1 ω̇1 − I1 (I2 − I3 )ω2 ω3 ω1 = 0
(54)
Multiplying by L1 = I1 ω1 gives
We have two similar equations for the other components. Adding them, the second terms
in these equations are seen to add to zero. The first terms add to
L1 (I1 ω̇1 ) + L2 (I2 ω̇2 ) + L3 (I3 ω̇3 ) = 0
(55)
I1 I˙1 + I2 I˙2 + I3 I˙3 = 0
(56)
d 1 2 2
[ (I I + I32 )] = 0
dt 2 1 2
(57)
This is
which is
d 2
L =0
dt
so the magnitude of the angular momentum remains unchanged.
(58)
(b) We have
I1 ω̇1 − (I2 − I3 )ω2 ω3 = 0
(59)
I1 ω̇1 ω1 − (I2 − I3 )ω2 ω3 ω1 = 0
(60)
Multiplying by ω1 gives
When we add the three equations, the second terms cancel. The first terms give
I1
1d 2
1d 2
1d 2
d
(ω1 ) + I2
(ω2 ) + I3
(ω3 ) = T = 0
2 dt
2 dt
2 dt
dt
6
(61)